1. Acids & Bases Calculating pH and Naming

2. Finding the pH of Solutions Self- ionization of water – the simple dissociation of water H 2 O H + + OH - Concentration of each ion in pure water: [H + ] = 1.0 x 10 -7 M + [OH - ] = 1.0 x 10 -7 M Where K w = 1.0 x 10 -14 Ion-product constant for water (K w ), Where K w = 1.0 x 10 -14 K w = [H + ] [OH - ] Acid [H + ] > [OH - ] Base [H + ] < [OH - ] Neutral [H + ] = [OH - ]

3. pH Scale pH = -log[H + ] pOH = -log[OH - ] pH + pOH = 14

4. [OH - ]pOHpH[H + ] 1 x 10 -14 1401 x 10 0 1 x 10 -13 1311 x 10 -1 1 x 10 -12 1221 x 10 -2 1 x 10 -11 1131 x 10 -3 1 x 10 -10 10Increasing acidity41 x 10 -4 1 x 10 -9 951 x 10 -5 1 x 10 -8 861 x 10 -6 1 x 10 -7 7Neutral71 x 10 -7 1 x 10 -6 681 x 10 -8 1 x 10 -5 591 x 10 -9 1 x 10 -4 4Increasing basicity101 x 10 -10 1 x 10 -3 3111 x 10 -11 1 x 10 -2 2121 x 10 -12 1 x 10 -1 1131 x 10 -13 1 x 10 0 0141 x 10 -14

5. Example If the [H + ] in a solution is 1.0 x 10 -5 M, is the solution acidic, basic or neutral? 1.0 x 10 -5 M What is the concentration of the [OH - ]? Use the ion-product constant for water (K w ): K w = [H + ] [OH - ] 1.0 x 10 -14 = [1.0 x 10 -5 ] [OH - ] 1.0 x 10 -14 = [OH - ] 1.0 x 10 -5 1.0 x 10 -(14-5) pH 5 = acidic 1.0 x 10 -9 OH -

6. Fill in the chart. [OH - ]pOHpH[H + ] 8 1x 10 -12 10 1 x 10 -3 5 1 × 10 -1 1.0 X 10 -8 1.0 X 10 -2 1.0 X 10 -6 6 212

7. Naming Acids Binary acids –Contains 2 different elements: H and another –Always has “hydro-” prefix –Root of other element’s name –Ending “-ic” –Examples: HI, H 2 S, HBr

8. Naming Acids Ternary Acids - Oxyacids –Contains 3 different elements: H, O, and another –No prefix –Name of polyatomic ion –Ending “–ic” for “-ate” and “–ous” for “- ite” –Examples: HClO 4, H 3 PO 4, HNO 2

9. Practice H 2 SO 3 –Sulfurous acid HF –Hydrofluoric acid H 2 Se –Hydroselenic acid Perchloric acid –HClO 4 Carbonic acid –H 2 CO 3 Hydrobromic acid –HBr